Question

[An Sơ Hạ]

Chứng minh các đẳng thức sau :

a) (sin x + cos x)² = 1 + 2sin x.cos x

b) sin⁴ x + cos⁴ x = 1 - 2sin² x.cos² x

c) tan² x - sin² x = tan² x.sin² x

d) sin⁶ x + cos⁶ x = 1 - 3sin² x.cos² x

e) sin x.cos x (1 + tan x)(1 + cot x) = 1 + 2sin x .cos x

Answer 1

[Akai Haruma]

a)

\((\sin x+\cos x)^2=\sin ^2x+2\sin x\cos x+\cos ^2x\)

\(=(\sin ^2x+\cos ^2x)+2\sin x\cos x=1+2\sin x\cos x\)

b)

\(\sin ^4x+\cos ^4x=\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x-2\sin ^2\cos ^2x\)

\(=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x\)

\(=1-2\sin ^2x\cos ^2x\)

c)

\(\tan ^2x-\sin ^2x=(\frac{\sin x}{\cos x})^2-\sin ^2x\)

\(=\sin ^2x\left(\frac{1}{\cos ^2x}-1\right)=\sin ^2x. \frac{1-\cos ^2x}{\cos ^2x}=\sin ^2x.\frac{\sin ^2x}{\cos ^2x}\)

\(=\sin ^2x\left(\frac{\sin x}{\cos x}\right)^2=\sin ^2x\tan ^2x\)

Answer 2

[Akai Haruma]

d)

\(\sin ^6x+\cos ^6x=(\sin ^2x)^3+(\cos ^2x)^3\)

\(=(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)\)

\(=\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x\)

\(=(\sin ^4x+\cos ^4x)-\sin ^2x\cos ^2x=1-2\sin ^2x\cos ^2x-\sin ^2x\cos ^2x\)

\(=1-3\sin ^2x\cos ^2x\) (theo kq phần b)

e)

\(\sin x\cos x(1+\tan x)(1+\cot x)=\sin x\cos x(1+\frac{\sin x}{\cos x})(1+\frac{\cos x}{\sin x})\)

\(=\sin x\cos x.\frac{\cos x+\sin x}{\cos x}.\frac{\sin x+\cos x}{\sin x}\)

\(=(\sin x+\cos x)^2=\sin ^2x+\cos ^2x+2\sin x\cos x\)

\(=1+2\sin x\cos x\)

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P/s: Nói chung cứ bám vào công thức \(\sin ^2x+\cos ^2x=1\)